CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Let APB be the tangent and take O as centre of the circle.

Let us suppose that MP{tex}\bot{/tex}AB does not pass through the centre.
Then,
{tex}\angle OPA = 90^\circ{/tex} [{tex}\because{/tex} Tangent is perpendicular to the radius of circle]
But {tex}\angle MPA = 90^\circ{/tex} [Given]
{tex}\therefore \angle OPA = \angle MPA{/tex}
This is only possible when point O and point M coincide with each other.
Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

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If2sinA=1then find the value of A

Posted by Shrwan Singh 7 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Ayush Kamal Kaushik 7 years, 1 month ago

Given that 2sin A=1 sinA =1/2 sin A=sin30 A=30

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Shakti Singh Bhati 5 years, 8 months ago

2sinA =1 sinA =1/2 sinA = sin30 A = 30

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If a and b are roots of equation x^2+ax-b=0 then find value of a and b

Posted by Hajira Tabassun 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Hemant Kumar 7 years, 1 month ago

Let p (x)= x^2 + ax - b =0 Now, a+b= - a/1 And a.b=-b/1 =~ a = -b / b =~a=-1 So... -1 +b = -(-1) b= 1+1 =2

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Its of trignometry

Posted by Noumi Thakur 7 years, 1 month ago

CBSE > Class 10 > Mathematics

0 answers

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Solve the quadratic equation 4√3 + 5x - 2√3 = 0

Posted by Shweta Sharma 7 years, 1 month ago

CBSE > Class 10 > Mathematics

3 answers

Hamza Hasan 7 years, 1 month ago

-2/5

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Chesta Pawan Manchanda 7 years, 1 month ago

X=-2√3/5

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Saket Choubey 7 years, 1 month ago

9548aba

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Find HCF of 144 &90

Posted by Parasappa Jaggal 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Yogita Ingle 7 years, 1 month ago

90 = 2 x 3 x 3 x 5 = 2 x 3² x 5
144 = 2 x 2 x 2 x 2 x 3 x 3 = 2⁴ x 3²
HCF (90, 144) = 2 x 3² = 18

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A toy is in the form of a cone mounted on a hemisphere of common base radious 7 cm. The total height of the toy is 31 cm. Find thr total surface area of the toy.

Posted by Abhishek Raikwar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Total surface area of toy = C.S.A of hemisphere + C.S.A. of a cone
{tex} = 2\pi {r^2} + \pi rl{/tex}
Here, r = 7cm, h = 24 cm
{tex}\therefore l = \sqrt {{r^2} + {h^2}} = \sqrt {{7^2} + {{24}^2}} = 25cm{/tex}
T.S.A. of toy {tex} = 2 \times \frac{{22}}{7} \times 7 \times 7 + \frac{{22}}{7} \times 7 \times 25{/tex}
= 308 + 550 = 858 cm²

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Show that the square of any positive integer is of the form 3m or,3m+1for some integer m

Posted by Khusi Gupta 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Anurag Yadav 7 years, 1 month ago

Let a be any positive integer

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A man walks a certain distance with certain speed.if he walks half km an hour faster,he takes 1hour less but he walks 1km an hour slower,he takes 3 more hours.find the distance covered by the man and speed of his walking.

Posted by Suriya Don 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Anirudh Mishra 7 years, 1 month ago

See in example of Ncert and rs argawal .

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Cos(15)=

Posted by Anirudh Mishra 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Shashwat Majge 7 years, 1 month ago

U can find it by using this formula: cos(A−B)=cos(A)cos(B)+sin(A)sin(B)
[Divide cos15 as cos[45-30]]

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35/6×50/7+11/2

Posted by Kushal Bains 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Abhijit Kshatri 7 years, 1 month ago

35/6×50/7+11/2 1750/42+11/2 Take LCM of 42 and 2 We get 42 So 1750+231 ____________ 42 =1981/42 =283/6

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What is the CSA of cylinder

Posted by Disha Rathore 7 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Sunil Hanotiya 7 years, 1 month ago

2πrh

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Sarath Sarath 7 years, 1 month ago

2 pie r.h

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Solution of exercise 9.1 Question no.10

Posted by Samriddhi Pal 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Check NCERT solutions here : https://mycbseguide.com/ncert-solutions.html

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Prove mid point theorem

Posted by Pradeep Kumar 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Let a line AB have two mid-points, say, C and D. Then
AB = AC + CB = 2AC . . . . (i) . . . [As C is the mid-point of AB]
and AB = AD + DB = 2AD . . . . (ii) [As D is the mid-point of AB]
From equation (i) and (ii)
AC = AD and CB = DB
But this will possible only when D lies on point C. So every line segment has one and only one mid-point.

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Sin theta= Cos (theta-6)

Posted by Viksita Bhardwaj 7 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Kunal J 7 years, 1 month ago

Given: sintheta = cos(theta-6) Now sintheta = cos(90- theta) So, cos(90- theta) = cos(theta-6) So, 90 - theta = theta-6 90 + 6 = theta + theta 96 = 2theta 96/2 = theta 48 = theta.

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Sreya Sreenivasan 7 years, 1 month ago

Sin theta=sin 6 Theta=6

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The radii of circular ends of a bucket of height 15 cm and 14 cm and r is greater than 14 cm. If volume of bucket is 5390 centimetre cube find the value of r

Posted by Purvi Kyal 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Given Height of the bucket (h) = 15 cm
r = 14 cm
R = ?
Now,
Volume of the bucket {tex}= \pi \times \frac { 1 } { 3 } \times \left( r ^ { 2 } + R ^ { 2 } + r R \right) \times h{/tex}
{tex}\Rightarrow 5390 = \frac { 22 } { 7 } \times \frac { 1 } { 3 } \times \left( 14 ^ { 2 } + R ^ { 2 } + 14 R \right) \times 15{/tex}
{tex}\Rightarrow 5390 = \frac { 110 } { 7 } \times \left( 196 + \mathrm { R } ^ { 2 } + 14 \mathrm { R } \right){/tex}
{tex}\Rightarrow \frac { 539 \times 7 } { 11 } = 196 + R ^ { 2 } + 14 R{/tex}
{tex}\Rightarrow{/tex} 343 = 196 + R² + 14R
{tex}\Rightarrow{/tex} R²+ 14R = 147
{tex}\Rightarrow{/tex} R² + 14R = 147 = 0
{tex}\Rightarrow{/tex} R²+ 21R - 7R -147 =0
{tex}\Rightarrow{/tex} R(R + 21) -7(R + 21) = 0
{tex}\Rightarrow{/tex} (R - 7)(R + 21) = 0
{tex}\Rightarrow{/tex} R = -21 or R = 7
{tex}\Rightarrow{/tex} R = 7 cm. ({tex}\because{/tex} R cannot be negative)

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x² - a² > 0 Then x < -a or x > a Explain why?

Posted by Saurav Kumar 7 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Saurav Kumar 7 years, 1 month ago

Then how x < -a ?

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Kunal J 7 years, 1 month ago

Since x^2 - a^2 > 0 So, x^2 > a^2 So, this implies x > a.

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X2+6x+5=0

Posted by Neeli Neeli Nasir 7 years, 1 month ago

CBSE > Class 10 > Mathematics

3 answers

Arohi . 7 years, 1 month ago

Sorry I have wrote it in different lines but it came together.try to understand it.

1Thank You

Arohi . 7 years, 1 month ago

x^2+5x+x+5, =x(x+5)+1(x+5), =(x+5)(x+1), x=-5, x=-1 Might help you:0

1Thank You

Neeli Neeli Nasir 7 years, 1 month ago

Factorization mefhod

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What is the value of sin-90

Posted by Prakriti Ghosh 7 years, 1 month ago

CBSE > Class 10 > Mathematics

1 answers

Vipul Sharma 7 years, 1 month ago

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O is the centreof the circle and BCD is a tangent to it at C. Prove that angle ABC + angle ACD= 90°

Posted by Naman Sen 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

{tex}\angle OCD = 90^\circ{/tex} (tangent and radii are {tex}\bot {/tex} to one another at the point of contact)
In {tex}\triangle{/tex}OCA,
OC = OA (radii of circle)
Hence, {tex}\angle OCA = \angle OAC{/tex} (angles opposite to equal sides are equal)
Also, {tex}\angle OCD = \angle OCA + \angle ACD{/tex}
{tex}90^\circ = \angle OAC + \angle ACD{/tex} {tex}\left( {\because \angle OCA = \angle OAC} \right){/tex}
{tex}90^\circ = \angle BAC + \angle ACD{/tex}
Hence, {tex}\angle BAC + \angle ACD = 90^\circ{/tex}
Hence proved.